Problem: Divide the following complex numbers. $ \dfrac{7+11i}{1+3i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${1-3i}$ $ \dfrac{7+11i}{1+3i} = \dfrac{7+11i}{1+3i} \cdot \dfrac{{1-3i}}{{1-3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(7+11i) \cdot (1-3i)} {(1+3i) \cdot (1-3i)} = \dfrac{(7+11i) \cdot (1-3i)} {1^2 - (3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(7+11i) \cdot (1-3i)} {(1)^2 - (3i)^2} = $ $ \dfrac{(7+11i) \cdot (1-3i)} {1 + 9} = $ $ \dfrac{(7+11i) \cdot (1-3i)} {10} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({7+11i}) \cdot ({1-3i})} {10} = $ $ \dfrac{{7} \cdot {1} + {11} \cdot {1 i} + {7} \cdot {-3 i} + {11} \cdot {-3 i^2}} {10} $ Evaluate each product of two numbers. $ \dfrac{7 + 11i - 21i - 33 i^2} {10} $ Finally, simplify the fraction. $ \dfrac{7 + 11i - 21i + 33} {10} = \dfrac{40 - 10i} {10} = 4-i $